# About Galois representation, which Galois extension should i use?

In the book "Rational Points on Elliptic Curves by J.Silverman and J.Tate" it is defined a representation $$\rho_n:Gal(\mathbb{Q}(E[n])/\mathbb{Q})\longrightarrow GL_2(\mathbb{Z}/n\mathbb{Z})\hspace{0.2cm}\forall n\geq 2$$

with $$\mathbb{Q}(E[n]):=\mathbb{Q}(x_1,y_1,...,x_{n^2-1},y_{n^2-1})$$, where $$E[n]=\{O,(x_1,y_1),...,(x_{n^2-1},y_{n^2-1})\}$$

But in other refferences i saw $$\rho_n:Gal(\overline{\mathbb{Q}}/\mathbb{Q})\longrightarrow GL_2(\mathbb{Z}/n\mathbb{Z})\hspace{0.2cm}\forall n\geq 2$$

I know $$\mathbb{Q}(E[n])\subset \overline{\mathbb{Q}}$$ beacuse $$\mathbb{Q}(E[n])/\mathbb{Q}$$ is algebraic, in fact, it is Galois.

Which Galois group should i use?

The map $$\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)\to\operatorname{GL}_2(\mathbb Z/n\mathbb Z)$$ factors through $$\operatorname{Gal}({\mathbb Q}(E[n])/\mathbb Q)\to\operatorname{GL}_2(\mathbb Z/n\mathbb Z)$$. So it's not really a matter of which Galois group to use, it's a matter of what you're doing. If $$n$$ is fixed, it's often easier to use the finite Galois group; but if you're going to take a sequence of increasing values of $$n$$, e.g., $$n=\ell^k$$ with $$k\to\infty$$, then you're better off using the Galois group of $$\overline{\mathbb Q}$$ so that you don't have to keep switching groups.
• Thank you very much!. I have a question about it. I saw $\overline{\mathbb{Q}}/\mathbb{Q}$ and $\mathbb{Q}(E[n])/\mathbb{Q}$ are Galois extension. Why is the Galois condition necessary in order to define representations? I think if $L/\mathbb{Q}$ is any extension and $\sigma:L\longrightarrow \overline{\mathbb{Q}}$ is a homomorphism of fields then $\sigma(x,y)=(\sigma(x),\sigma(y))$ define an action over $E[n]$ without Galois condition. Commented May 8, 2020 at 19:22
• @danihelovick A (linear) representation, by definition, is a group homomorphism from a group $G$ to a linear group GL$(V)$. So you can't really call it a representation. OTOH, sure, $\sigma$ will define a group action, as long as $E[n]\subset E(L)$. But if not, then $\sigma$ does not give a well-defined map from $E[n]$ to itself. Commented May 8, 2020 at 20:51